Conditional probability of correlated gaussians Let $(X,Y)$ be a 2D Gaussian, with non-zero correlation. Let $I$ be an interval centered around the mean of $X$, and $J$ an interval centered around the mean of $Y$. I want to show
$$
P(X \in I | Y \in J) \geq  P(X \in I),
$$
i.e. that the conditional probability is greater than or equal to the unconditional one.
I am sure this is true thanks to simulation, but I cannot find a proof of this result. It feels very natural that, for two correlated gaussians, knowing that one is closer to its mean implies that the probability that the other is also closer to its mean is higher. I tried using the conditional density of $X | Y=a$ and integrating over $a$, but I couldn't find a minoration.
 A: I'm going to switch your variables in the statement so that the discussion uses them in a more conventional sense of studying $Y$ conditional on the values of $X$.  So, the objective is to show that
$$\Pr(Y\in J\mid X\in I) \ge \Pr(Y\in J).\tag{1}$$
Think of this in terms of regression:


*

*We will exploit the fact that the distribution of $Y$ conditional on $X=x$ is a standard distribution $F$ (in this case Normal with mean zero and some given variance $\sigma^2$) whose location has been shifted to $f(x)$ for some function $f$ (the regression function).  By letting $J-f(x)$ stand for all values $y-f(x)$ where $y\in J$,this means $$\Pr(Y\in J\mid X=x) = {\Pr}(Y-f(x)\in J-f(x)\mid X=x)=\int_{J-f(x)} \mathrm{d}F(y).\tag{2}$$

*Let's suppose that intervals  centered around $0$ are maximum-probability intervals for $F$: among all intervals of the same width, these have the largest probability.  That's clearly the case for symmetric unimodal distributions like the Normal.  This can be written $$\int_J \mathrm{d}F(y) \ge \int_{J-a} \mathrm{d}F(y)\tag{3}$$ for any number $a$.
Assume $\Pr(X\in I) \gt 0$. (The other case is trivial to prove).  This allows us to write 
$$\Pr(Y\in J\mid X\in I) = \frac{\Pr(Y\in J\text{ and } X\in I)}{\Pr(X\in I)}.\tag{4}$$
Now, letting $G$ stand for the distribution of $X$, we can compare the two sides of $(1)$ by means of $(2)$, applying $(3)$ with $a=f(x)$, and simplifying the resulting double integral:
$$\eqalign{
\Pr(Y\in J\text{ and } X\in I) &= \int_I \Pr(Y\in J\mid X=x) \mathrm{d}G(x) \\
&=\int_I\int_{J-f(x)}\mathrm{d}F(y)\ \mathrm{d}G(x) \\
&\le \int_I\int_J \mathrm{d}F(y)\mathrm{d}G(x) \\
&= \int_J \mathrm{d}F(y) \int_I\mathrm{d}G(x) \\
&=\Pr(Y\in J)\Pr(X\in I).
}$$
In light of $(4)$, dividing both sides by $\Pr(X\in I)$ produces $(1)$, QED.
A: You want to prove that for any $x^\star \geq 0$ and $y^\star \geq 0$
we have $$ \text{Pr}\{ |X| \leq x^\star,\, |Y| \leq y^\star \} \geq
\text{Pr}\{ |X| \leq x^\star\} \text{Pr}\{ |Y| \leq y^\star \} $$
which is a famous result known as Sidak inequality. It holds for an
arbitrary centered elliptically contoured bivariate distribution,
including the Gaussian.
For a quite general formulation, see this paper by das
Gupta et al.
where more useful results are given.
A: Your desired result is a special case of the recently-proven Gaussian correlation inequality.
This may seem like overkill (and as the other answers demonstrate, it is), but the proof of the more general result is relatively short and available on the arXiv.
